A polylog competitive algorithm for the k-server problem Nikhil Bansal (IBM) Niv Buchbinder (Open Univ.) Aleksander Madry (MIT) Seffi Naor (Technion)

The k-server Problem k servers lie in an n-point metric space. Requests arrive at metric points. To serve request: Need to move some server there. Goal: Minimize total distance traveled. Objective: Competitive ratio Move Nearest Algorithm

The Paging/Caching Problem Set of pages {1,2,…,n}, cache of size k<n. Request sequence of pages 1, 6, 4, 1, 4, 7, 6, 1, 3, … a) If requested page already in cache, no penalty. b) Else, cache miss. Need to fetch page in cache (possibly) evicting some other page. Goal: Minimize the number of cache misses. K-server on the uniform metric. Server on location p = page p in cache 1 n...

Previous Results: Paging Paging (Deterministic) [Sleator Tarjan 85]: Any deterministic algorithm >= k-competitive. LRU is k-competitive (also other algorithms) Paging (Randomized): Rand. Marking O(log k) [Fiat, Karp, Luby, McGeoch, Sleator, Young 91]. Lower bound H k [Fiat et al. 91], tight results known.

K-server conjecture [Manasse-McGeoch-Sleator ’88]: There exists k competitive algorithm on any metric space. Initially no f(k) guarantee. Fiat-Rababi-Ravid’90: exp(k log k) … Koutsoupias-Papadimitriou’95: 2k-1 Chrobak-Larmore’91: k for trees.

Randomized k-server Conjecture There is an O(log k) competitive algorithm for any metric. Uniform Metric: log k Polylog for very special cases (uniform-like) Line: n 2/3 [Csaba-Lodha’06] exp(O(log n) 1/2 ) [Bansal-Buchbinder-Naor’10] Depth 2-tree: No o(k) guarantee

Our Result Thm: There is an O(log 2 k log 3 n) competitive* algorithm for k-server on any metric with n points. * Hiding some log log n terms Key Idea: Multiplicative Updates

Our Approach Hierarchically Separated Trees (HSTs) [Bartal 96]. Any Metric Allocation Problem (uniform metrics): [Cote-Meyerson-Poplawski’08] (decides how to distribute servers among children) O(log n) Allocation instances K-server on HST

Outline Introduction Allocation Problem Fractional view of Randomized Algorithms Fractional Caching Algorithm What makes Allocation Problem harder? The Fix

Allocation Problem Uniform Metric At each time t, request arrives at some location i Request = (h t (0),…,h t (k)) [monotone: h(0) ¸ h(1) … ¸ h(k)] Upon seeing request, can reallocate servers Hit cost = h t (k i ) [k i : number of servers at i] Total cost = Hit cost + Move cost Eg: Paging = cost vectors ( 1,0,0,…,0) *Total servers k(t) can also change (let’s ignore this)

Allocation to k-server Thm [Cote-Poplawski-Meyerson]: An online algorithm for allocation s.t. for any  > 0, i) Hit Cost (Alg) · (1+  ) OPT ii) Move Cost (Alg) ·  OPT gives ¼ O(d  1/d)) competitive k-server alg. on depth d HSTs d = log (aspect ratio) So,  = poly(1/  ) polylog(k,n) suffices *HSTs need some well-separatedness *Later, we do tricks to remove dependence on aspect ratio We do not know how to obtain such an algorithm.

Outline Introduction Allocation Problem Fractional view of Randomized Algorithms Fractional Caching Algorithm What makes Allocation Problem harder? The Fix

Fractional View of Randomized Algorithms To specify a randomized algorithm: i) Prob. distribution on states at time t. ii) How it changes at time t+1. Fractional view: Just specify some marginals. Eg. Paging, actual algorithm = distribution over k-tuples but, Fractional: p 1,…,p n s.t. p 1 + …+ p n = k Cost: If p 1,…,p n changes to q 1,…,q n,pay (1/2)  i |p i – q i | Not too weak: Fractional Paging -> Randomized Paging (2x loss)

Fractional Allocation Problem x i,j : prob. of having j servers at location i (at time t)  j x i,j = 1 (prob. distribution)  i  j j x i,j · k (global server bound) Cost: Hit cost with h(0),…,h(k) =  j x i,j h(j) Moving  mass from (i,j) to (i,j’) costs  |j’-j| Surprisingly, fractional allocation does not give good randomized alg. for allocation problem.

A gap example Allocation Problem on 2 points Requests alternate on locations. Left: (1,1,…,1,0) Right: (1,0,…,0,0) Any integral solution must pay  (T) in T steps. Claim: Fractional Algorithm pays only T/(2k). X L,0 = 1/k x L,k = 1-1/k X R,1 = 1 No move cost. Hit cost of 1/k on left requests. LeftRight

Fractional Algorithm Suffices Thm (Analog of Cote et al): Suffices to have fractional allocation algorithm with (1+ ,  (  )) guarantee. Gives a fractional k-server algorithm on HST Thm (Rounding): Fractional k-server alg. on HSTs -> Randomized Alg. with O(1) loss. Thm (Frac. Allocation): Design a fractional allocation algorithm with  = O(log (k/  )).

Outline Introduction Allocation Problem Fractional view of Randomized Algorithms Fractional Caching Algorithm What makes Allocation Problem harder? The Fix

Fractional Paging Algorithm Current state, p 1,…,p n s.t.  i p i =k. Say request at 1 arrives. Algorithm: Need to bring 1-p 1 mass into p 1. Rule: For each page i decrease p i / 1–p i +  (  = 1/k) Intuition: If p i close to 1, be more conservative in evicting. Multiplicative Update: d(1-p) / (1-p) 0 Pg 1Pg 2 Pg n … p1p1 1-p 1 1-p 2 p2p p n pnpn

Potential Fn. Analysis Will Show: On(t) +  (t) -  (t-1) · O(log k) Off(t) Contribution of page i to  : 0 if p i =0, log(k+1) if p i =1.  =0 if online and offline state coincide. Suppose, page 1 is requested. Analyze in two steps: First offline moves, then online If offline moves a server  increases by · log(k+1) The claim holds easily.  = 1/k

Analysis Will show: On(t) +  (t) -  (t-1) · O(log k) Off(t) Suppose  mass enters p 1. p i decreases by dp i =  (1-p i +  ) / N Offline Online On = {i : p i > 0} Key Obs: For each page i in On\Off  decreases by dp i ¢ d  / dp i = dp i (1/(1-p i +  )) =  /N At least |On|-k such pages, so potential drop >=  (|On|-k)/N Claim: Potential drop ¸  Pf: N =  i 2 On (1-p i +  ) = |On|-k +  |On| ¼ |On|-k.

What makes Allocation Harder? Paging: If location 1 requested, Know that OPT also has server on 1. Allocation Problem: Not so clear. Say, already 10.5 servers at location 1. Should we add even more? May be OPT just has 1 server there.

An Instructive Example Case 1: K servers, and allocation problem on 2k locations. Cost Vectors = (1,0,…,0) At Locations: 1, 2, …, 2k 1, 2,…, 2k, … Case 2: 2 k servers in total. 1/2 server on locations 1,…,k-1 and k+1/2 on location k. Vector (1,0,…,0) on location 1,2,…,k-1 and vector (1,1,…,1,0,…,0) on k. Right thing: Release the servers on location k. Much better solution: 1 server on each location k... k k+1

The Fix Suppose cost vector  j = ( , ,…, ,0,…,0) at location 1. (i.e. cost  if · j servers, 0 otherwise) Hit cost Y=  (x 1,0 + …+ x 1,j ) Increase servers by ¼ Y Fix number: For each location i (including 1), rebalance prob. mass by multiplicative update. (location 1) 012 jj+1 k Recall  j x ij = 1, 8 i …… Each x i,j (except last j) increases / x i,j

Proof Idea Eg: Location i contributes 3 log (1+k) to . Key observation: For every cut · j, x i, · j increases / x i, · j. Lemma: For any offline state,  pays for online movement. Pf: Each cut for j > k * i, decreases potential by  /|ON-k|. There are |ON-k| such cuts. Location i OPT ON

Concluding Remarks Removing dependence on aspect ratio. HST -> Weighted HST with O(log n) depth. Extend Allocation to weighted star. Main question: Can we remove dependence on n. 1. Metric -> HST 2. But even on HST (lose depth of HST)

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